Work Done by a Field Along a Helical Path — Conservative-Field Shortcut: 04-BS-1, December 2016, Q7

Question 7: Work Done by a Field Along a Helical Path — Conservative-Field Shortcut 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A vector field $\mathbf F=(x^2,y,-z)$ and a helical path from $(0,2,0)$ (at $t=0$) to $(3\pi,0,2)$ (at $t=\pi/2$; check: $x=6(\pi/2)=3\pi$, $y=2\cos(\pi/2)=0$, $z=2\sin(\pi/2)=2$ ✓).

Find. The work $W=\displaystyle\int_C\mathbf F\cdot d\mathbf r$.

Approach. Check whether $\mathbf F$ is conservative before attempting the (messy) direct parametrized line integral. If it is, the work reduces to a potential difference between the endpoints, independent of the actual path taken.

Test for conservativeness. $\mathbf F=(x^2,y,-z)$ has $\operatorname{curl}\mathbf F=\mathbf 0$ identically (each component depends only on its own variable, so all mixed partials in the curl vanish). Equivalently, a potential $\varphi$ exists directly: $\varphi_x=x^2\Rightarrow\varphi=\tfrac{x^3}{3}+h(y,z)$; then $\varphi_y=y\Rightarrow h=\tfrac{y^2}{2}+k(z)$; then $\varphi_z=-z\Rightarrow k=-\tfrac{z^2}{2}$. So $$\varphi(x,y,z)=\frac{x^3}{3}+\frac{y^2}{2}-\frac{z^2}{2}.$$
Evaluate the potential at the endpoints. $$\varphi(3\pi,0,2)=\frac{(3\pi)^3}{3}+0-\frac{4}{2}=\frac{27\pi^3}{3}-2=9\pi^3-2,\qquad \varphi(0,2,0)=0+\frac{4}{2}-0=2.$$
Work as a potential difference. $$W=\varphi(\text{end})-\varphi(\text{start})=(9\pi^3-2)-2=\boxed{9\pi^3-4}\approx275.9$$ (Cross-check: directly parametrizing $\mathbf F\cdot d\mathbf r$ along the given helix and integrating $t=0\to\pi/2$ reproduces $9\pi^3-4$ exactly, confirming the path-independence shortcut.)

Quantity	Result
Potential $\varphi(x,y,z)$	$\tfrac{x^3}{3}+\tfrac{y^2}{2}-\tfrac{z^2}{2}$
$\varphi$ at start $(0,2,0)$	$2$
$\varphi$ at end $(3\pi,0,2)$	$9\pi^3-2$
Work $W$	$9\pi^3-4\approx275.9$

Concept & Theory

Topic. Conservative vector fields and the fundamental theorem for line integrals.

Key relations. $\mathbf F=\nabla\varphi$ for some scalar potential $\varphi$ iff $\operatorname{curl}\mathbf F=\mathbf 0$ (on a simply connected domain). When this holds, $\displaystyle\int_C\mathbf F\cdot d\mathbf r=\varphi(\text{end})-\varphi(\text{start})$ for any path $C$ between the two points.

Why this works. Here $\mathbf F$ is "separable" — each component is a function of only its own coordinate — which is a quick visual cue (though not a proof) that a potential likely exists; confirming $\operatorname{curl}\mathbf F=\mathbf 0$ (or, as done here, directly constructing $\varphi$ successfully) verifies it rigorously. Once conservative, the specific helical shape of the path becomes irrelevant to the answer — only the endpoints matter.

Common pitfall. Not checking conservativeness first and instead grinding through the full trigonometric line integral (here involving $\cos^2t,\sin^2t$ terms and a $t^2$-type term from $x^2=36t^2$) — still correct if done carefully, but far more error-prone; forgetting to verify the given path's start/end parameter values actually correspond to the stated points before computing $\varphi$ there.

Source. Stewart, Calculus: Early Transcendentals, Ch. 16 (§16.3 The Fundamental Theorem for Line Integrals).

Question 7 of 8: Work Done by a Field Along a Helical Path — Conservative-Field Shortcut

Question 7: Work Done by a Field Along a Helical Path — Conservative-Field Shortcut 20 marks